If I have the following map:
$\phi_s: \mathbb Z_{p^a} \rightarrow \Bbb Z_{p^b}$ defined by multiplication by $kn/d$ where $n = p^b$ and $d = \gcd(p^a, p^b)$ and $k= p^s$ and $0 \leq k < d$ and I know that if $a\geq b$ and we are not in the case of the zero homomorphism i.e. we took $s = b -1$, we will have:
1- ker $\phi_{s} = Z_{p^{a-s+1}}.$
2- coker $\phi_s = Z_{p^{a-s}}.$
How can I start proving the above results about the kernel and cokernel, should I use the properties of SES or what? could anyone help me please?
Unfortunately, the formula you have is not right. Consider the case where $a=b$ and $s=0$. Then $n=d$ and $k=1$, so the map you've defined is the identity map and hence has trivial kernel and cokernel, but your formula would give its kernel to be $\mathbb{Z}/p^{a+1}$ and its cokernel to be $\mathbb{Z}/p^a$.
If you don't mind, I'll use the notation $\phi_s$ for what you refer to by $k\Lambda$. So $\phi_s$ takes $\overline{1}\in\mathbb{Z}/p^a$ to $\overline{p^{s+b}/d}\in\mathbb{Z}/p^b$, where $d=\gcd(p^a,p^b)$. Note that the only positive divisors of $p^a$ and $p^b$ are powers of $p$, so in fact $d=p^{\min\{a,b\}}$, and we can write $\phi_s(\overline{1})=\overline{p^{s+b-\min\{a,b\}}}$. To see that $\phi_s$ is a well-defined map, you need to show that $p^a\phi_s(\overline{1})=\overline{0}\in\mathbb{Z}/p^b$. This amounts to showing that $p^ap^{s+b-{\min\{a,b\}}}$ is divisible by $p^b$ in $\mathbb{Z}$, ie that $a+s+b-\min\{a,b\}\geqslant b$. Since $s\geqslant 0$, this is clear.
Now, $\overline{n}\in\ker\phi_s$ if and only if $\overline{0}=\phi_s(\overline{n})=\overline{n{p^{s+b-\min\{a,b\}}}}\in\mathbb{Z}/p^b$, which is in turn true if and only if $np^{s+b-\min\{a,b\}}$ is divisible by $p^b$ in $\mathbb{Z}$. If $p^k$ is the highest power of $p$ dividing $n$, then this is true if and only if $k+s+b-\min\{a,b\}\geqslant b$, ie if and only if $k+s\geqslant\min\{a,b\}$. You are assuming that $\phi_s$ is not the zero map, so we know $s<\min\{a,b\}$. Thus a generator for $\ker\phi_s$ is given by $\overline{p^{\min\{a,b\}-s}}$. The order of $\overline{p^{\min\{a,b\}-s}}$ is $p^{a-(\min\{a,b\}-s)}=p^{a+s-\min\{a,b\}}$, and subgroups of cyclic groups are cyclic, so $\fbox{$\ker\phi_s\cong\mathbb{Z}/p^{a+s-\min\{a,b\}}$}$. In particular, in the case you brought up, where $b\leqslant a$ and $s=b-1$, we have $\ker\phi_s\cong\mathbb{Z}/p^{a-1}$.
Now, the cokernel of $\phi_s$ is the quotient of $\mathbb{Z}/p^b$ by the image of $\phi_s$. Quotients of cyclic groups are cyclic, and cyclic groups are uniquely determined by their cardinalities, so the cokernel of $\phi_s$ will be determined by its size. Now, by the first isomorphism theorem, the image of $\phi_s$ is isomorphic to the quotient $\frac{\mathbb{Z}/p^a}{\ker\phi_a}$. Since $|\mathbb{Z}/p^a|=p^a$, and $|\ker\phi_s|=p^{a+s-\min\{a,b\}}$ by the computation above, we have $$|\operatorname{im}\phi_s|=\frac{p^a}{p^{a+s-\min\{a,b\}}}=p^{\min\{a,b\}-s}.$$ (Again, you are assuming $s<\min\{a,b\}$, so this number makes sense.) Since $|\mathbb{Z}/p^b|=p^b$, we can hence compute $$|\operatorname{coker}(\phi_s)|=\frac{|\mathbb{Z}/p^b|}{|\operatorname{im}\phi_s|}=\frac{p^b}{p^{\min\{a,b\}-s}}=p^{b+s-\min\{a,b\}},$$ and so $\fbox{$\operatorname{coker}\phi_s\cong\mathbb{Z}/p^{b+s-\min\{a,b\}}$}$. In particular, in the case you brought up, where $b\leqslant a$ and $s=b-1$, we have $\operatorname{coker}\phi_s\cong\mathbb{Z}/p^{b-1}$.